# Infinitely repeated game: prisoners' dilemma

I have done most of the legwork, but I have fallen short at the final hurdle. Could you please correct my mistake(s)?

Given the condition $x>0$ I need to consider the infinitely repeated game where the above strategic form is the stage game. The discount factor: $\delta=\frac{1}{2}$.

I need to find a further condition on the game, such that we have a subgame perfect equilibrium in which both players "cooperate" in each period.

What I have done so far

The profile of strictly dominated strategies $(u,r)$ constitutes the cooperative strategy where $x>0$ and $x \ne 1$.

The prescription (cooperative strategy) will therefore be $(u,r)=(x+1,x+1)$

The grim trigger is $(d,l) = (x,x)$

Payoff from obeying the prescription:

$(x+1) + \delta(x+1) + \delta^2(x+1) +\delta^3(x+1)+... = \frac{x+1}{1-\delta}$

Payoff from deviating (here comes the mistake):

$2x + \delta x + \delta^2 x +\delta^3 x +... =\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$

The part in red should apparently be:

$$\frac{(2-\delta)x}{1-\delta}$$

I don't understand why I am wrong, could you please explain? My understanding was that if a player deviates in period 1 in order to obtain $2x$ then both players would have to play the grim trigger profile $(d,l)$ from period 2 to perpetuity. As a result, the players would be missing out on $\delta(x+1)$. The payoff numerator would therefore be $\color{red}{2x - \delta(x+1)}$ instead of $\color{blue}{(2-\delta)x}$.

Given that the correct numerator is $\color{blue}{(2-\delta)x}$, I can easily finish the question:

$$\frac{x+1}{1-\delta} \ge \frac{(2-\delta)x}{1-\delta}$$

Therefore, we would have a subgame perfect equilibrium iff $x \le 2$.

The problem is that my mistake lies in:

$2x + \delta x + \delta^2 x +\delta^3 x +...$ $=\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$

But, I'm not sure why. I would like to know why I am wrong and why $\color{blue}{(2-\delta)x}$ is the correct numerator.

Thanks.

the payoff from play the trigger strategy will be: $$\sum_{i=0}^{\infty}(x+1) \delta^i=\frac{x+1}{1-\delta}$$
if I deviate and I play $l$ or $d$ the payoff will be $$2x + \sum_{i=1}^{\infty}x \delta^i = 2x+ x\frac{\delta}{1-\delta}=\frac{2x(1-\delta) + x\delta}{1-\delta}= \frac{x(2-\delta)}{1-\delta}$$
then, the condition is $$\frac{x+1}{1-\delta} \geq \frac{x(2-\delta)}{1-\delta}$$